If  $a,bc\in R(a\ne 0)anda+2b+4c=0$ then equation  $a{x}^2+bx+c=0$ has_________

$$\begin{gathered} x=a+2b+4c=0 \\ \frac{a}{4}+\frac{b}{2}+c=0 \\ a.{\left(\frac{1}{2}\right)}^2+b.\left(\frac{1}{2}\right)+c=0 \end{gathered}$$

It is clean that one roots  $\frac{1}{2}$
$$\begin{gathered} \alpha +\frac{1}{2}=-\frac{b}{a} \\ \alpha =-\frac{1}{2}-\frac{b}{a} \end{gathered}$$ 
Which depends upon a & b